214 Chapter 6: Distributions of Sampling Statistics
wherex=
∑n
i= 1 xi/n. It follows from this identity that
(n−1)S^2 =
∑n
i= 1
Xi^2 −nX^2
Taking expectations of both sides of the preceding yields, upon using the fact that for any
random variableW,E[W^2 ]=Var(W)+(E[W])^2 ,
(n−1)E[S^2 ]=E
[ n
∑
i= 1
Xi^2
]
−nE[X^2 ]
=nE[X 12 ]−nE[X^2 ]
=nVar(X 1 )+n(E[X 1 ])^2 −nVar(X)−n(E[X])^2
=nσ^2 +nμ^2 −n(σ^2 /n)−nμ^2
=(n−1)σ^2
or
E[S^2 ]=σ^2
That is, the expected value of the sample variance S^2 is equal to the population
varianceσ^2.
6.5Sampling Distributions from a Normal Population
LetX 1 ,X 2 ,...,Xnbe a sample from a normal population having meanμand varianceσ^2.
That is, they are independent andXi∼N(μ,σ^2 ),i=1,...,n. Also let
X=
∑n
i= 1
Xi/n
and
S^2 =
∑n
i= 1
(Xi−X)^2
n− 1
denote the sample mean and sample variance, respectively. We would like to compute
their distributions.